From a point 100 m above …

Let AB be the surface of the lake and P be the point of observation such that {tex}AP = 100 m{/tex}. Let C be the position of the helicopter and C' be its reflection in the lake.
Then, {tex}CB = C'B.{/tex}

Let PM be perpendicular from P on CB.Then, {tex}\angle{/tex}{tex}CPM = 30 ^\circ{/tex}and {tex}\angle{/tex}C'PM = {tex}60^\circ{/tex}
Then, {tex}CM = h, \ CB = h + 100.{/tex}
In right {tex}\triangle{/tex}CMP
{tex}\tan 30^{\circ}=\frac{\mathrm{C} \mathrm{M}}{\mathrm{PM}} \Rightarrow{\frac{1}{ \sqrt{3}}}=\frac{h}{\mathrm{PM}}{/tex}
{tex}\Rightarrow PM =\sqrt3 h{/tex}...(i)
In right {tex}\triangle{/tex}PMC'
{tex}\tan 60^{\circ}=\frac{\mathrm{C}^{\prime} \mathrm{M}}{\mathrm{PM}} \Rightarrow \sqrt{3}=\frac{\mathrm{C}^{\prime} \mathrm{B}+\mathrm{BM}}{\mathrm{PM}}{/tex}
{tex}\Rightarrow \sqrt{3}=\frac{h+100+100}{P M}{/tex}
{tex}\Rightarrow P M=\frac{h+200}{\sqrt{3}}{/tex} ...(ii)
From (i) and (ii), we get 
{tex} \sqrt{3} h=\frac{h+200}{\sqrt{3}}{/tex}
{tex}  3h = h + 200 \\  2h = 200\\   h = 100{/tex}
Now,
{tex}CB = CM + MB\\ = h + 100\\ = 100 + 100\\ = 200\\{/tex} 
Hence, the height of the helicopter from the surface of the lake = 200 m